# Angular momentum per unit energy in the Schwarzschild spacetime

I am reading Chapter 9 of Hartle's Gravity book, and I'm completely stumped on the following thing. In equation 9.45 he claims that for circular orbits in a Schwarschild geometry:

$$\frac{l}{e}=\left(Mr\right)^{1/2}\left(1-\frac{2M}{r}\right)^{-1}\tag{9.45}$$ And my question is how the above equation comes about.

l and e are derived from the killing vectors:

$$e=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}$$

$$l=r^2 \sin^2\theta \frac{d\phi}{d\tau}$$

He hints that this equation (9.45) comes from the fact that the energy equals the effective potential:

$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$

and that the radius of the orbit must correspond to the one that minimizes the effective potential:

$$r_{\rm min}=\frac{l^2}{2M}\left[1+\sqrt{1-12\left(\frac{M}{l}\right)^2}\right]$$

I get these last two equations, but I am lost on how they can be used to derive his 9.45.

• I don't have a copy of Hartle, but have you tried writing down the geodesic equation and finding $d\phi/\der t$ for a circular orbit?
– user4552
Commented Apr 24, 2019 at 23:48

Once you have introduce $$e$$ and $$l$$ the remaining equation of motion is

$$e^2 = \left(\frac{dr}{d\tau}\right)^2 +V(r)$$

with

$$V(r) = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$.

Being a circular orbit requires two things. First of all the radius must not change, i.e. $$\frac{dr}{d\tau}=0$$. Second, this condition must be maintained over time meaning that $$\frac{d^2r}{d\tau^2}=0$$ as well.

The first condition simply implies,

$$e^2 = V(r)$$.

The second can be obtained from the first equation above by taking the time derivative and applying the chain rule which leads to

$$0 = -2 \frac{d^2r}{d\tau^2} = V'(r) = \frac{M}{r^2} - 2 l^2\frac{r-3M}{r^4}$$.

That is, $$r$$ must be at the minimum of the the potential $$V(r)$$. Now instead of solving this equation for $$r_{\rm min}$$, as you did, solve it to find $$l$$,

$$l^2 = M\frac{r^2}{r-3M}$$.

Plugging this into the equation for $$e^2$$ gives

$$e^2 = \frac{(r-2M)^2}{r(r-3M)}$$.

The rest should be obvious.